/**
 * Copyright (c) 2020-2024, huli.com
 * All rights reserved.
 */
package com.xiaohujr.xuexue.leetcode.meituan;

/**
 * @author Xue Xue (xuexue1@huli.com)
 * @version 1.0
 * @since 2021/2/23
 */
public class FindMedianSortedArrays {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int m = nums1.length;
        int n = nums2.length;
        int left = (m + n + 1) / 2;
        int right = (m + n + 2) / 2;
        return (findKth(nums1, 0, nums2, 0, left) + findKth(nums1, 0, nums2, 0, right)) / 2.0;
    }

    //i: nums1的起始位置 j: nums2的起始位置
    public int findKth(int[] nums1, int i, int[] nums2, int j, int k) {
        if (i >= nums1.length) {
            return nums2[j + k - 1];
        }//nums1为空数组
        if (j >= nums2.length) {
            return nums1[i + k - 1];
        }//nums2为空数组
        if (k == 1) {
            return Math.min(nums1[i], nums2[j]);
        }
        int midVal1 = (i + k / 2 - 1 < nums1.length) ? nums1[i + k / 2 - 1] : Integer.MAX_VALUE;
        int midVal2 = (j + k / 2 - 1 < nums2.length) ? nums2[j + k / 2 - 1] : Integer.MAX_VALUE;
        if (midVal1 < midVal2) {
            return findKth(nums1, i + k / 2, nums2, j, k - k / 2);
        } else {
            return findKth(nums1, i, nums2, j + k / 2, k - k / 2);
        }
    }

    public static void main(String[] args) {
        new FindMedianSortedArrays().findMedianSortedArrays(new int[]{1, 3}, new int[]{2});
    }

}
